mt_sn - EE 278 Wednesday, November 11, 2009 Statistical...

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Unformatted text preview: EE 278 Wednesday, November 11, 2009 Statistical Signal Processing Handout #16 Midterm Examination Solutions 1. Inequalities. a. e S is a convex function so by Jensens inequality E( e S ) e E( s ) = e Since E( S ) = 1. b. S is a concave function so by Jensens inequality E( S ) E( S ) 1 2 = 1 Since E( S ) = 1. 2. Exponentials. a. To find E( S | S > 5), we need to find the pdf of S given the event S > 5. Let event A = { S > 5 } . f S | A ( s ) = f S ( s ) P ( S A ) if s A , and 0 , otherwise = e- s P ( S > 5) = e- s integraltext 5 e- s ds = e- s e- 5 = e- ( s- 5) if s > 5, and 0 , otherwise Let s prime = s- 5 so s = s prime + 5. E( S | A ) = integraldisplay 5 sf S | A ( s ) ds = integraldisplay 5 se- ( s- 5) ds = integraldisplay ( s prime + 5) e- s prime ds prime = integraldisplay s prime e- s prime ds prime + integraldisplay 5 e- s prime ds prime = 1 + 5 = 6 b. To find E( S | S + T ), we first should recognize that E( S + T | S + T ) = E( S | S + T ) + E( T | S + T ) S + T = 2 E( S | S + T ) Page 2 of 4 EE 278, Fall 2009 The first line follows from linearity of expectation. The second line follows because S and T are independent and have the same distribution so E( S | S + T ) = E( T | S + T )....
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This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.

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mt_sn - EE 278 Wednesday, November 11, 2009 Statistical...

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